Get key from a list of values in dictionary?

Question

Let's say I have:

ref = ['<var>', '<id>', '<expr>']
val = [['a', 'b', 'c'], 'a', '1+1']
dicio = dict(zip(ref, val))

now, I know that by doing

list(dicio.keys())[list(dicio.values()).index('a')]

It returns <id>. But let's say that you only had one value associated per key, so

val = [['a', 'b', 'c'], 'b', '1+1']

How could I get <var> without listing ['a', 'b', 'c']? Thank you.

@BrokenBenchmark most of the bidirectional dictionary solutions won't work when the value is a list, since lists aren't hashable. — Barmar, Jun 20 '22 at 17:09
@Barmar Then the list can be transformed into a tuple instead. — BrokenBenchmark, Jun 20 '22 at 17:10
@barmar Also, could a recursive solution be implemented here to traverse the values list and find the proper value? — Cardstdani, Jun 20 '22 at 17:18
If you don't want to list `['a', 'b', 'c']` what do you want to use instead? — Barmar, Jun 20 '22 at 17:20
@Barmar Precisely what Cardstdani suggested. But I'm not sure how to implement this. — Vitor L., Jun 20 '22 at 17:32
https://stackoverflow.com/questions/35117847/python-how-to-search-a-nested-list-using-recursion/68152990#68152990 — Barmar, Jun 20 '22 at 17:35

Cardstdani · Answer 1 · 2022-06-20T17:16:47.280

Just replace "a" value by the list ['a', 'b', 'c']:

print(list(dicio.keys())[list(dicio.values()).index(['a', 'b', 'c'])])

You will get as output the value associated with the index you inserted before:

<var>

Or you can use list comprehension to traverse the list of dict values:

ref = ['<var>', '<id>', '<expr>']
val = [['a', 'b', 'c'], 'a', '1+1']
dicio = dict(zip(ref, val))

selectedVal = list(i for i in list(dicio.values()) if "a" in i)[0]
print(list(dicio.keys())[val.index(selectedVal)])

Which outputs the same as the previous solution.

Get key from a list of values in dictionary?

1 Answers1